Amplitude of a block oscillating on a spring

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SUMMARY

The discussion focuses on calculating the amplitude of a block oscillating on a spring with a mass of 0.25 kg and a spring constant of 200 N/m. The block is initially stretched to 0.15 m and given a speed of 3.0 m/s. The correct amplitude is determined to be 0.18 m, which is derived from energy considerations rather than initial conditions. The key equations utilized include the angular frequency (omega) and the relationship between maximum speed and amplitude.

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  • Understanding of harmonic motion and oscillation principles
  • Familiarity with Hooke's Law and spring constants
  • Knowledge of energy conservation in mechanical systems
  • Ability to apply formulas for angular frequency and maximum speed in oscillatory motion
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  • Study the concept of energy conservation in oscillatory systems
  • Learn how to derive amplitude from initial conditions and energy equations
  • Explore the relationship between angular frequency and oscillation parameters
  • Investigate the effects of damping on oscillatory motion
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for examples of spring dynamics and energy conservation principles.

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Homework Statement


a .25kg block oscillates on the end of a spring with a spring constant of 200N/m. If the oscillation is started by elongating spring to 0.15m and giving the block a speed of 3.0m/s. The amplitude is?


Homework Equations


omega= sqrt(k/m), Vm = (omega)Xm


The Attempt at a Solution



omega= sqrt (200/.25) = 28.28 (3m/s) = 28.28Xm (3/28.28) = .106

The correct answer is .18m..what am i missing?
 
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The spring is stretched by 0.15 m and the block has 3 m/s speed at the same time. 0.15 is not the maximum displacement and 3 m/s is not the maximum speed. Calculate the energy, and get the maximum speed from it.

ehild
 

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